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  "body": "\n## Statement\n\nLet $ABC$ be a [[triangle|0CTNFGYMM2]] in the Euclidean space $\\mathbb{R}^n$, and let $d$ be the [[Euclidean distance|0NPPGHQNXQ]]. Let $D$ be a point of the segment $BC$, and let $\\angle$ denote the [[angle at a point|0GJAJC6TNZ]].\nIf $\\angle BAD = \\angle DAC$, then\n$$\\frac{d(B, D)}{d(D, C)} = \\frac{d(A, B)}{d(A, C)}.$$\nConversely, if $D$ is a point of the segment $BC$ with this equality, then $\\angle BAD = \\angle DAC$.\n\n## Remarks\n\nIn words, the bisector of the angle at $A$ cuts the opposite side in the ratio of the two adjacent sides.\nIn an isosceles triangle with $d(A, B) = d(A, C)$, the bisector from $A$ meets $BC$ at its midpoint.\nA similar statement holds for the bisector of the exterior angle at $A$, which meets the line $BC$ outside the segment.\n\n## History\n\nThe theorem is Proposition 3 of Book VI of Euclid's Elements.\n\n## Explanations\n\n### L1 human\n\nTake the triangle with $d(A, B) = 6$ and $d(A, C) = 3$. The bisector of the angle at $A$ cuts $BC$ into two parts. The part next to $B$ is twice as long as the part next to $C$.\n",
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